Homogeneous linear equations of order n with constant coefficients

As happen with the linear systems of order 1, one ODE of order $n$ has $n$ linearly independent solutions in such a way that any solution to an homogeneous ODE will be a linear combination of these linearly independent solutions. Therefore, to solve the ODE will consist in finding these $n$ functions.

The ODE we consider is $a_{n} \cdot y^{(n)} (x) + a_{n - 1} \cdot y^{(n - 1)} + \dots + a_{1} \cdot y^{'} + a_{0} \cdot y = 0$ where $a_{i}$ are constants.

Example

An example of a honogenous ODE of order $n$ would be: $y^{″} + y = 0$

Then we define the characteristical polynomial of the ODE as: $a_{n} \cdot λ^{n} + a_{n - 1} \cdot λ^{n - 1} + \dots + a_{1} \cdot λ + a_{0} = 0$ and we look at its $n$ roots.

The characteristical polynomial is easy to write, it is just necessary to change $y$ for $λ$ and to raise it to the corresponding order of the derivative.

Example

For example, in the ODE that we have given, the associated characteristic polynomial is: $λ^{2} + 1 = 0$ .

This polynomial has two combined complex roots: $λ_{1} = i, λ_{2} = - i$

Then

If $λ$ are real and simple the solution is of the form: $e^{λ x}$
If $λ$ are real of multiplicity m the solutions are of the form: $e^{λ x}, x \cdot e^{λ x}, x^{2} \cdot e^{λ x}, \dots, x^{m - 1} \cdot e^{λ x}$
If $λ = a + b i$ are complex and simple, the two solutions are of the form: $e^{a x} \cos (b x), e^{a x} \sin (b x)$ (there are two solutions because whenever a complex root exists the conjugate also appears)
If $λ = a + b i$ are complex of multiplicity m, the two solutions are of the form: $e^{a x} \cos (b x), x \cdot e^{a x} \cos (b x), \dots, x^{m - 1} e^{a x} \cos (b x) e^{a x} \sin (b x), x \cdot e^{a x} \sin (b x), \dots, x^{m - 1} e^{a x} \sin (b x)$

Then, once we find these $n$ solutions, the general solution of the ODE will be a linear combination of these $n$ solutions.

Example

Let's come back to the example at the beginning. As our polynomial took two (simple) combined complex roots we are in case 3. Therefore the solution is: $y (x) = c_{1} \cdot \cos x + c_{2} \cdot \sin x$ where the constants will be determined by the initial conditions (in the event that we would know them).